PII. S0161171204401124 http://ijmms.hindawi.com © Hindawi Publishing Corp.
MULTIVALENT FUNCTIONS AND
Q
KSPACES
HASI WULAN
Received 13 January 2004
We give a criterion for q-valent analytic functions in the unit disk to belong to QK, a Möbius-invariant space of functions analytic in the unit disk in the plane for a nonde-creasing functionK:[0,∞)→[0,∞), and we show by an example that our condition is sharp. As corollaries, classical results on univalent functions, the Bloch space, BMOA, and
Qpspaces are obtained.
2000 Mathematics Subject Classification: 30D35, 30D45.
1. Introduction. For analytic univalent functionfin the unit disk∆, Pommerenke [8] proved thatf∈Ꮾif and only iff∈BMOA, which easily implies a result of Baernstein II [4] about univalent Bloch functions: if g(z)≠0 is an analytic univalent function in ∆, thenlogg∈BMOA. We know that Pommerenke’s result mentioned above was generalized toQpspaces for allp, 0< p <∞, by Aulaskari et al. (cf. [2, Theorem 6.1]). Their result can be stated as follows.
Theorem1.1. Let f be an analytic function in∆such that
|w−w0|<1
n(w, f )dA(w)≤A <∞, (1.1)
for allw0∈C, wheren(w, f )denotes the number of roots of the equationf (z)=win ∆counted according to their multiplicity anddA(z)is the Euclidean area element on∆. Thenf∈Ꮾ(Ꮾ0)if and only iff∈Qp(Qp,0)for allp∈(0,∞).
Here,Qpand its subspaceQp,0, 0< p <∞, denote the spaces of analytic functions f in∆defined, respectively, as follows (cf. [1,3]):
Qp=
f:f analytic in∆,sup
a∈∆
∆
f(z)2g(z, a)pdA(z) <∞
,
Qp,0=
f∈Qp: lim |a|→1
∆
f(z)2g(z, a)pdA(z)=0
,
(1.2)
whereg(z, a)=log 1/|ϕa(z)|is a Green’s function in∆with pole ata∈∆, andϕa(z)= (a−z)/(1−az)¯ is a Möbius transformation of∆.
is defined as follows:
Ꮾ=f:fanalytic in∆,fᏮ=sup
z∈∆
1−|z|2f(z)<∞. (1.3)
Similar to the above we have Q1,0=VMOA, the space of all analytic functions of vanishing mean oscillation (cf. [5]), andQp,0=Ꮾ0for allp∈(1,∞), whereᏮ0denotes the little Bloch space defined by
Ꮾ0=
f∈Ꮾ: lim |z|→1
1−|z|2f(z)=0. (1.4)
In the present paper, we consider a more general spaceQK (see below) and show that all the above-mentioned results are true for spaceQK. Our contribution gives an extended version of Pommerenke’s theorem, which is also a slight improvement of all the above results, and the proof presented here is independently developed.
LetK:[0,∞)→[0,∞)be a right-continuous and nondecreasing function. Recall that the spaceQKconsists of analytic functionsf in∆for which
f2 QK=sup
a∈∆
∆
f(z)2Kg(z, a)dA(z) <∞; (1.5)
f∈QKbelongs to the spaceQK,0if
∆
f(z)2Kg(z, a)dA(z) →0, |a|→1. (1.6)
Modulo constants,QKis a Banach space under the norm defined in (1.5). It is clear that
QK is Möbius-invariant and a subspace of the Bloch spaceᏮ(cf. [6]). For 0< p <∞,
K(t)=tpgives the spaceQp. ChoosingK(t)=1, we get the Dirichlet spaceᏰ.
By [6, Proposition 2.1] we know that if the integral 1/e
0
K log1
ρ
ρ dρ=
∞
1
K(t)e−2tdt (1.7)
is divergent, then the spaceQKis trivial; that is, the spaceQK contains only constant functions. From now on, we assume that the function K : [0,∞)→[0,∞) is right-continuous and nondecreasing and that the integral (1.7) is convergent. Without loss of generality, we can assume thatK(1) >0. For a general theory forQKspaces, see [6,11].
2. Main results. A functionf analytic in the unit disk is said to beq-valent if the equationf (z)=whas never more thanqsolutions. Let
p(ρ)=21π
2π
0 n
ρeiφ, fdφ. (2.1)
If
R
0
or
p(R)≤q, R >0, (2.3)
where q is a positive number, we say thatf is areally mean q-valent or circumfer-entially meanq-valent, respectively (cf. [7, pages 38 and 144]). It is clear that iff is circumferentially meanq-valent, thenfis areally meanq-valent.
Note that if (1.1) holds,fwill be areally meanq-valent in∆for someq >0. We know that iffis univalent, thenfmust be areally and circumferentially mean 1-valent. Thus, it is natural to conjecture that Pommerenke’s result andTheorem 1.1are also true for the areally and circumferentially meanq-valent functions.
We know that the spaceQKcan be nontrivial ifKis not too big at infinity (see condi-tion (1.7)). For such functionsK, the properties ofQKdepend essentially on the behavior ofKnear the origin. From [6, Theorems 2.3 and 2.5], we know thatQK=Ꮾ(QK,0=Ꮾ0) if and only if
1
0
1−r2−2K log1 r
r dr <∞. (2.4)
A natural idea is to look for an integral condition which is weaker than that given by (2.4) such thatf∈Ꮾ(Ꮾ0)if and only iff∈QK(QK,0)for some specialf. For the areally meanq-valent case, we present the main result in this paper as follows.
Theorem2.1. Letf be an areally meanq-valent function in∆. If
1
0
log 1 1−r
2
(1−r )−1K log1
r
r dr <∞, (2.5)
then
(i) f∈Ꮾif and only iff∈QK;
(ii) f∈Ꮾ0if and only iff∈QK,0.
Note that (2.4) implies (2.5) since(log 1/(1−r ))2≤4e−2/(1−r )for 0< r <1, but the converse is not true. For example,K(t)=tgives that (2.5) holds but (2.4) fails. By [6, Theorems 2.3 and 2.5], (2.5) is also necessary forTheorem 2.1(i) and (ii) in casef is an areally meanq-valent function in∆.
In the light of the following example it is impossible to drop the assumption of areally meanq-valence of the functionsfinTheorem 2.1. Indeed, chooseK1(t)=t2α−1and
f1(z)= ∞
j=1
2−j(1−α)z2j, 1
2< α <1. (2.6)
Forr∈[3/4,1), we findkso that 1/2≤2k(1−r ) <1. Using the inequality logr≥
2(r−1), 1/2< r <1, we see that
2π
0
f1
r eiθ2dθ=2π ∞
j=1
2j2αr2j+1−2
≥2π (1−r )−2α ∞
j=1
2j(1−r )2α
exp−2j+2(1−r )
≥2−2α+1π (1−r )−2α ∞
j=1
2(j−k)(2α)exp−2j−k+2
≥2−2α+1π (1−r )−2α ∞
j=0
2j2αexp−2j+2
=C(α)(1−r )−2α.
(2.7)
Hence
sup
a∈∆
∆
f 1(z)
2
K1g(z, a)dA(z)
≥
∆
f1(z)2K1 log 1
|z|
dA(z)
=
1
0
K log1
r
r dr
2π
0
f1
r eiθ2dθ
≥C(α)
1
3/4(1−r )
−2α log1 r
2α−1 r dr .
(2.8)
Since the last integral is divergent, we conclude thatf1∉QK.
Theorem2.2. Letfbe a circumferentially meanq-valent and nonvanishing function
in∆. If (2.5) holds, thenlogf∈QK.
It is clear that the integral in (2.5) is convergent forK(t)=tp,p >0. Thus, we have
the following result which extendsTheorem 1.1.
Corollary2.3. Let f be an areally meanq-valent function in∆,0< p <∞. Then (i) f∈Ꮾif and only iff∈Qp;
(ii) f∈Ꮾ0if and only iff∈Qp,0.
3. Proofs. In the proofs of Theorems2.1and2.2, we need two lemmas, the first one can be considered as a generalization of a result of Pommerenke (cf. [9, page 174]).
Lemma3.1. Letf be areally meanq-valent in∆. Then
2π
0
fr eiθ2
dθ≤4qπ
M√r , f2
1−r ,
1
2< r <1, (3.1)
Proof. If 1/2< r <1, we obtain
|z|<√r
f(z)2
dA(z)=
√r
0 ρ
2π
0
fρeiθ2 dθ dρ
≥1
4(1−r ) 2π
0
fr eiθ2 dθ.
(3.2)
Sincef is areally meanq-valent, we deduce that
2π
0
fr eiθ2dθ≤ 4
1−r
|z|<√r
f(z)2dA(z)
≤1−4r
|w|<M(√r ,f )n(w, f )dA(w)
≤4qπ
M√r , f2
1−r ,
(3.3)
which provesLemma 3.1.
Lemma3.2. LetKbe defined as inSection 1. Then
(i) QK,0⊂Ꮾ0;
(ii) an analytic functionfbelongs toᏮ0if and only if there exists anr∈(0,1)such that
lim |a|→1
∆(a,r )
f(z)2Kg(z, a)dA(z)=0, (3.4)
where∆(a, r )= {z∈∆:|ϕa(z)|< r}.
Proof. See [6, Thereom 2.4].
Now we turn to give the proofs of our main theorems.
Proof ofTheorem2.1. We first prove (i). SinceQK⊂Ꮾ, it suffices to prove that if a Bloch functionf is areally meanq-valent in∆, thenf∈QK. We use the change of variablew=ϕa(z)to deduce that
∆\∆(a,1/2)
f
(z)2Kg(z, a)dA(z)
=
∆\∆(a,1/2)
f (z)−f (a)2K log 1
ϕa(z)
dA(z)
=
1/2<|w|<1
f◦ϕa(w)−f (a)2
K log 1
|w|
dA(w)
=
1
1/2
K log1
r
r
2π
0
f◦ϕar eiθ−f (a)2 dθ dr .
It is known that ifg∈Ꮾ, then
g(z)−g(0)≤1
2gᏮlog 1+|z|
1−|z|. (3.6)
Choosingg=f◦ϕa−f (a)and observing thatgᏮ= fᏮ, we obtain
Mr , f◦ϕa−f (a)≤12fᏮlog1+r
1−r. (3.7)
It follows from (3.5) andLemma 3.1that
∆\∆(a,1/2)
f(z)2Kg(z, a)dA(z)
=
1
1/2K log 1
r
r
2π
0
(f◦ϕar eiθ−f (a)2dθdr
≤4qπ
1
1/2
K log1
r
M√r , f◦ϕa−f (a)2(1−r )−1r dr
≤qπ Cf2 Ꮾ
1
1/2
K log1
r log
1 1−r
2
(1−r )−1r dr .
(3.8)
On the other hand, we have
∆(a,1/2)
f(z)2Kg(z, a)dA(z)
≤ f2 Ꮾ
∆(a,1/2)
1−|z|2−2
Kg(z, a)dA(z)
= f2 Ꮾ
∆(0,1/2)
1−|w|2−2K log 1
|w|
dA(w)
≤4πf2 Ꮾ
1/2
0
K log1
r
r dr .
(3.9)
Combining the upper bounds given by (3.8), (3.9), and (2.5), we see thatf∈QK, which proves part (i) ofTheorem 2.1.
To prove (ii), we assume thatf is an areally mean q-valent function in ∆which is also inᏮ0. ByLemma 3.2(i), it suffices to prove thatf∈QK,0. ByLemma 3.2(ii), there exists anr0, 1/2< r0<1, such that
lim |a|→1
∆(a,r0)
f(z)2
Kg(z, a)dA(z)=0. (3.10)
Now we show that
lim |a|→1
∆\∆(a,r0)
f(z)2
By the proof of part (i) and assumption (2.5), we see that
∆\∆(a,r0)
f
(z)2Kg(z, a)dA(z)
=
1
r0
K log1
r
r
2π
0
f◦ϕa
r eiθ−f (a)2dθ dr
≤4qπ
1
r0
K log1
r
M√r , f◦ϕa−f (a)2(1−r )−1r dr
≤qπf2 Ꮾ
1
r0
K log1
r log
1+r
1−r
2
(1−r )−1r dr <∞
(3.12)
for alla∈∆. Thus, for any givenε >0, there exists anr1,r0< r1<1, such that
1
r1
K log1
r
M√r , f◦ϕa−f (a)2(1−r )−1r dr < ε (3.13)
for alla∈∆. Hence, what we need to prove is that
lim |a|→1
r1
r0
K log1
r
M√r , f◦ϕa−f (a)2(1−r )−1r dr=0. (3.14)
In fact, we have r1
r0
K log1
r
M√r , f◦ϕa−f (a)2(1−r )−1r dr
≤Cr0, r1K log 1 r0
Mr2, f◦ϕa−f (a)2,
(3.15)
wherer2=√r1andC(r0, r1)is a constant depending onr0andr1. Defineft(z)=f (tz) for 0< t <1 and then
Mr2, f◦ϕa−f (a)
2
≤2
1
4f−ft 2 Ꮾ log
1+r2 1−r2
2
+Mr2, ft◦ϕa−ft(a)2
. (3.16)
Sincef∈Ꮾ0,f−ftᏮ→0,t→1. Also,
max |z|≤r2
ft◦ϕa(z)−ft(a)≤1−|a|2 1−r2
2max |w|≤t
f(w), (3.17)
which implies that
lim |a|→1M
r2, ft◦ϕa−ft(a)
Thus we have (3.14). Hence
lim |a|→1
∆
f(z)2Kg(z, a)dA(z)=0, (3.19)
which shows thatf∈QK,0. The proof ofTheorem 2.1is complete.
Proof ofTheorem2.2. Assume thatfis a nonvanishing circumferentially mean
q-valent function in ∆. According to [7, Theorem 5.1], we have logf ∈ Ꮾ. From [7, Lemma 5.2] and the argument in the beginning of the proof of [7, Theorem 5.1], we see that we can define a single-valued branch off (z)1/qwhich is circumferentially
mean 1-valent in∆and such that on each circle{|w| =R}there exists a point which is not assumed byf (z)1/q. It follows that
∞
−∞n
logρ+iφ,1 qlogf
dφ=
2π
0 n
ρeiφ, f1/qdφ≤2π ,
|w|<R
n(w,logf )dA(w)≤4π Rq,
(3.20)
which means that logfis areally meanq1-valued in∆for someq1>0. It follows from Theorem 2.1that logf∈QK.
4. Further discussion. In [10] we studied the conditions for analytic univalent Bloch functionf to belong toQKspaces. The log-order of the functionK(r )is defined as
ρ=lim
r→∞
log+log+K(r )
logr , (4.1)
where log+x=max{logx,0}, and if 0< ρ <∞, the log-type of the function K(r )is defined as
σ=lim
r→∞
log+K(r )
rρ . (4.2)
Theorem4.1. Letfbe an analytic univalent function in∆and letK:[0,∞)→[0,∞)
satisfy thatK(t)=O(tlog 1/tp)ast→0for somep >0. If the log-orderρand the log-typeσ ofKsatisfy one of the conditions
(i) 0≤ρ <1,
(ii) ρ=1andσ <2,
thenf∈Ꮾif and only iff∈QK.
We note thatTheorem 4.1can be viewed as a consequence ofTheorem 2.1. In fact, conditions (i) and (ii) ofTheorem 4.1 show that the spaceQK is not trivial. That is, the integral (1.7) is convergent in this case. Suppose thatK(t)=O((tlog 1/t)p),t→0.
There exist anr0∈(1/2,1)and a constantC >0 such that both log 1/r≤2(1−r )and
K log1
r
≤C
log1
rlog log
1
r
−1p
hold forr0< r <1. Thus
1
0 log 1 1−r
2
(1−r )−1K log1
r
r dr
=
r0
0 +
1
r0
log 1 1−r
2
(1−r )−1K log1
r
r dr
≤ log 1 1−r0
2 1−r0
−1r0 0 K log
1
r
r dr
+C
1
r0
log 1 1−r
2
(1−r )−1
log1
rlog log
1
r
−1p r dr
≤C1+C2
1
r0
log 1 1−r
2+p
(1−r )p−1r dr
≤C1+C2
∞
R0
e−pss2+pds
≤C1+C2p−3−pΓ(3+p) <∞.
(4.4)
For a general analytic functionf, we have the following theorem.
Theorem4.2. Suppose that (2.5) holds. If
sup
a∈∆
|z|<r
f◦ϕa(z)2dA(z)=O log 1 1−r
2
, (4.5)
then
(i) f∈Ꮾif and only iff∈QK;
(ii) f∈Ꮾ0if and only iff∈QK,0.
Proof. We know that
2π
0
f◦ϕa
r eiθ2dθ≤1−4r
|z|<√r
f◦ϕa(z)2 dA(z)
≤1−1rO log 1 1−√r
2
≤1C−r log 1 1−r
2 .
(4.6)
The proof can be completed by an argument similar to that used in the proof of
Theorem 2.1.
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Hasi Wulan: Department of Mathematics, Shantou University, Shantou, Guangdong 515063, China
